self-entry-level test
TRANSCRIPT
Self-Entry-Level Test
Institute of Economics
Institute for Statistics and Econometrics
Kiel Universtiy, Germany
Dear Candidate,
the following questions are not an examination in the usual sense, but merely a voluntary
test. The main purpose of this test is to help YOU to find out whether you might be a suc-
cessful participant in our master’s degree programme. The knowledge required to answer these
questions - without much looking-up and without any (!) learning - is what we expect you to
bring with you, as part of the skills acquired in your bachelor studies.
In the application process for any of our master’s degree programmes, we can only evaluate you
based on your grades and credit points. We cannot take into account your answers to these
questions since we do not know the amount of help you might have used. But if you fail to
answer 40% of the test questions easily and decide to come to Kiel anyway, the probability of
you leaving the programme after the first examinations without success is increased - wasting
valuable time and money.
So, please be honest to yourself when answering the following questions!
Questions of the following sections are relevant for the single programmes:
Economics:
Mathematics, Econometrics, Statistics, Microeconomics, and Macroeconomics
Quantitative Economics:
Mathematics, Econometrics, Statistics, Microeconomics, and Macroeconomics
Quantitative Finance:
Mathematics, Econometrics, and Statistics
Environmental and Resource Economics:
Mathematics, Econometrics, and Microeconomics
1
Mathematical Questions
1. Determine the solution to the following integrals:
(a)∫
2x + ex/8 + cos(2x)dx
(b)∫ √
4x−2/3
+ 4x4dx, x > 0
(c)∫ 10
1exp(− ln(a
x))dx, a > 0
(d)∫ b
01
exp(2x)dx, b > 0
(e)∫ π
0cos(x)dx
2. Determine the solution to the following integrals by partial integration:
(a)∫ 1
1/2ln(4x2)dx
(b)∫x2θ exp(−θx)dx, θ > 0, x > 0
3. Obtain the first and second order derivatives for the following functions at t=0:
(a) f(t) = (pet + 1− p)n, n ∈ N, p ∈ (0, 1)
(b) f(t) = exp(µt+ 12σ2t2), σ > 0
4. Obtain the first order derivatives for the following functions:
(a) f(t) = ln( 3√
(t− a)(t+ a)), |t| > |a|
(b) f(t) = 1(t−1)a
, t > 1
(c) f(t) = cos( 4√t), t > 0
(d) f(t) = sin(cos(t))
(e) f(t) = ln(ln(t)), t > 1
5. Find the limit for n ∈ IN or x ∈ IR, if it exists:
(a) limn→∞
( (12
)n+ 2 · 4n
5 · 4n − 2 · 3n
)
(b) limn→∞
(3√n4 + n2 − 1
(n− 2√n)2 + 2
)
(c) limn→∞
(−1)n2n
2n3 − n2
(d) (using de l’Hopital’s rule) limx→∞
x2e−x
6. Let f(x) = 21−x.
(a) Find the derivative of f(x) up to order 3.
2
(b) Give the n-th derivative f (n)(x) for n ∈ IN0.
(c) Find the Taylor polynomial T 2f (x) of order 2 in x0 = 0.
(d) Give the Taylor series T∞f (x) in x0 = 0.
7. Consider the following Cobb-Douglas production function as given:
f(K,L) = aKbLc, a, b, c > 0;K > 0, L > 0
The partial derivatives are given by f ′K(K,L) = abKb−1Lc and f ′L(K,L) = acKbLc−1.
(a) Calculate the partial differential dfK with respect to K in (K0, L0)=(1,4).
(b) Calculate the total differential df in (K0, L0) = (1, 4) for c = 12.
8. Use the Lagrangian method to determine the critical points of f(x, y) = x2 + y2 subject
to the constraint y = 1x, x > 0.
9. Use the Lagrangian method to solve the critical points of the Cobb-Douglas utility
function U(x, y) = axby1−b, x, y, a > 0, 0 < b < 1 subject to the budget restriction
g(x, y) = 2x+ 3y = 10.
10. Use the Lagrangian method to determine the critical points of the utility function U(x1, x2) =
(x1−1)0.4(x2−2)0.8, x1 > 1, x2 > 2 subject to the budget constraint g(x1, x2) = x1+2x2 =
11.
11. Consider f(x, y) = exp(− 2x
+ y) with x 6= 0.
(a) Determine the first and second order partial derivatives.
(b) Calculate the total differential df in (x0, y0) = (2, 1).
12. Consider f(x, y) = x2 + y2 + 4 and g(x, y) = 2x3 + 2y3 + x2 + y2 with x, y 6= 0. Use
the Lagrangian method to solve for the critical points of f subject to the constraint that
g(x, y) = 0.
13. Calculate the determinants of the following matrices:
(a) A =
1 −3 4
−4 1 3
2 −2 3
(b) B =
1 −3 4
3 2 0
3 1 2
3
(c) C =
2a −1 1
3a 0 3
0 1 1
(d) D =
−1 2 3 4
2 0 0 4
1 1 2 1
4 3 −1 7
14. Determine the eigenvalues of the following matrices:
(a) A =
(0 1
4 0
)
(b) B =
(2 1
1 2
)
(c) C =
7 0 0
0 6 0
0 0 8
(d) D =
0 1 2
0 0 3
0 0 0
(e) E =
2 −2 3
0 3 −2
0 −1 3
15. Let
A =
(1 2
2 4
), B =
(4 −6
−2 3
), C =
(2 1
3 2
),
D =
(−2 7
5 −1
), E =
(1 1
).
Perform the following operations, if possible:
(a) A ·A
(b) A ·B
(c) A ·C
(d) A ·D
(e) ET ·A · E
(f) E ·A · ET
4
16. Find the inverses of the following matrices:
(a) A =
(1 −3
−3 5
)
(b) B =
1 3 2
3 5 0
0 5 7
(c) C =
1 2 4
8 6 3
8 6 2
5
Mathematical Questions: RESULTS
1. (a) = 2x 1ln(2)
+ 8ex/8 + 0.5 sin(2x) + c
(b) = 0.9449x2/3 − 43x−3 + c
(c) = 992a
(d) = 0.5(1− e−2b)
(e) = 0
2. (a) = 0.3863
(b) = −e−θx(x2 + 2xθ
+ 2θ2
)
3. (a) f ′(t)|t=0 = np, f ′′(t)|t=0 = n2p2 + np(1− p)
(b) f ′(t)|t=0 = µ, f ′′(t)|t=0 = µ2 + σ2
4. (a) = 2t3(t2−a2)
(b) = −a(t− 1)−a−1
(c) = 2 sin( 4√t)t−3/2
(d) = − cos(cos(t)) sin(t)
(e) = 1t ln(t)
5. (a) = 0.4
(b) = 1
(c) = 0
(d) = 0
6. (a) f ′(x) = −21−x ln(2), f ′′(x) = 21−x(ln(2))2, f ′′′(x) = −21−x(ln(2))3
(b) = 21−x(− ln(2))n
(c) = 2− 2 ln(2) + (ln(2))2
(d) =∑∞
k=0(− ln(2)x)k
k!
7. (a) = ab4c∆K
(b) = 2ab∆K + a4∆L
8. x = y = ±1, λ = ±2
9. x = 5b, y = 103
(1− b)
10. x1 = 3, x2 = 4, λ = 0.4595
6
11. (a) f ′x(x, y) = 2f(x, y)x−2, f ′y(x, y) = f(x, y)
f ′′xx(x, y) = 4f(x, y)(x−4 − x−3), f ′′yy(x, y) = f(x, y), f ′′xy(x, y) = f ′x(x, y)
(b) = 0.5∆x+ ∆y
12. x = y = −0.5, λ = 2
13. (a) = −21
(b) = 10
(c) = 0
(d) = 108
14. (a) λ1,2 = ±2
(b) λ1 = 3, λ2 = 1
(c) λ1 = 8, λ2 = 7, λ3 = 6
(d) λ1,2,3 = 0
(e) λ1 = 2, λ2,3 = 3±√
2
15. (a) =
(5 10
10 20
)
(b) =
(0 0
0 0
)
(c) =
(8 5
16 10
)
(d) =
(8 5
16 10
)(e) not defined
(f) = 9
16. (a) = −14
(5 3
3 1
)
(b) =
17.5 −5.5 −5
−10.5 3.5 3
7.5 −2.5 −2
(c) =
−0.6 2 −1.8
0.8 −3 2.9
0.0 1 −1.0
7
Econometric Questions
1. What is a cumulative distribution function, a probability distribution function (for dis-
crete random variables), and a probability density function (for continuous random vari-
ables)?
2. What is an expected value, a variance and a standard deviation of a random variable Y ?
3. Characterize a joint and a conditional probability distributions.
4. What is the normal distribution, the t distribution, the F distribution, and the χ2 distri-
bution?
5. What means convergence in probability and convergence in distribution?
6. What is simple random sampling? What means that the observations Y1, ..., Yn of a
random sample are independently and identically distributed (i.i.d.)?
7. How is the sample average Y computed?
8. Show that if Y1, ..., Yn are i.i.d. with mean µY and variance σ2Y then:
(a) Y is an unbiased and consistent estimator of the population mean,
(b) the sampling distribution of Y has mean µY and variance σ2Y
= σ2Y /n,
(c) the law of large numbers says that Y converges in probability to µY , and
(d) the central limit theorem says that the standardized version of Y ,(Y − µY )/σY , has
a standard normal distribution when n is large.
9. What is the difference between population and sample regression line?
10. What is this: β = (X′X)−1X′y?
11. Explain the use of PX = X(X′X)−1X′ and MX = In −PX.
12. Show that if (1) the regression errors, ui, have a mean of zero conditional on the regres-
sors Xi, (2) the sample observations are i.i.d. random draws from the population, (3)
the regressors are not linearly dependent, and (4) large outliers are unlikely, then the
OLS estimators of the population model Yi = Xiβ + ui are unbiased, consistent, and
asymptotically normally distributed.
13. What is the purpose and meaning of R2, adjusted R2, and the standard error of the
regression (SER)?
14. Show that the squared SER can be written a s2u = U′MXU
n−k−1.
8
15. How can single and joint hypotheses be tested using t and F -statistics? What is the
p-value?
16. Construct a 95% confidence interval for a regression coefficient.
17. What is the difference between traditional and heteroscedasticity-consistent standard er-
rors?
18. State the Gauss-Markov theorem.
19. Show that if the assumptions of question 12 hold and the regression errors are additionally
homoscedastic and normally distributed, then:
(a) U|X ∼ N (0n, σ2uIn).
(b) β|X ∼ N (β, σ2u(X
′X)−1).
(c) n−k−1σ2u× σ2
u ∼ χ2n−k−1, and
(d) tj =βj−βj
σu√
[(X′X)−1]jj∼ t(n− k − 1)
20. Why can multicollinearity matter?
21. What is an omitted variable bias?
22. Explain other forms of regressor endogeneity such as the choice of an incorrect functional
form, measurement error, and simultaneous causality.
23. Interpret the regression coefficients for different functional forms: linear-linear, log-linear,
linear-log, and log-log.
24. What is the use of interaction terms?
9
Econometric Questions: RESULTS
1. The cumulative probability distribution function of a random variable evaluated
at a particular value is the probability that the random variable is less than or equal to
that particular value.
The probability distribution of discrete random variables is the list of all possible
values of the variable and the probability that each value will occur. These probabilities
sum to 1.
Because a continuous random variable can take on a continuum of possible values, the
probability distribution used for discrete variables, which lists the probability of each
possible value of the random variable, is not suitable for continuous variables. Instead,
the probability is summarized by the probability density function. The area under the
probability density function between any two points is the probability that the random
variable falls between those two points.
2. The expected value of a random variable Y , denoted E(Y ), is the long-run average value
of the random variable over many repeated trials or occurrences. The expected value of
a discrete random variable is computed as a weighted average of the possible outcomes
of that random variable, where the weights are the probabilities of that outcome. The
expected value of Y is also called the expectation of Y or the mean of Y and is denoted
µY .
The variance and standard deviation measure the dispersion or the “spread” of a probabil-
ity distribution. The variance of a random variable Y , denoted var(Y ), is the expected
value of the square of the deviation of Y from its mean: var(Y ) = E[(Y −µY )2]. Because
the variance involves the square of Y , the units of the variance are the units of the square
of Y , which makes the variance awkward to interpret. It is therefore common to measure
the spread by the standard deviation, which is the square root of the variance and is
denoted σY . The standard deviation has the same unit as Y .
3. The joint probability distribution of two discrete random variables, say X and Y , is
the probability that the random variables simultaneously take on certain values, say x and
y. The probabilities of all possible (x, y) combinations sum to 1. The joint probability
distribution can be written as the function Pr(X = x, Y = y).
The distribution of a random variable Y conditional on another random variable X taking
on a specific value is called the conditional distribution of Y given X. The condi-
tional probability that Y takes on the value y when X takes on the value x is written
Pr(Y = y|X = x).
10
4. A continuous random variable with a normal distribution has the familiar bell-shaped
probability density. The function defining the normal probability density is given by:
fY (y) =1
σ√
2πe−
12
( y−µσ
)2
The normal density with mean µ and variance σ2 is symmetric around its mean and
has 95% of its probability between µ − 1.96σ and µ + 1.96σ. The Normal distribution
with mean µ and variance σ2 is expressed concisely as N(µ, σ2). The standard normal
distribution is the normal distribution with mean µ = 0 and variance σ2 = 1 and is
denoted by N(0, 1).
The Student t distribution with m degrees of freedom is defined to be the distribution
of the ratio of a standard normal random variable, divided by the square root of an
independently distributed chi-squared random variable with m degrees of freedom divided
by m.
The F distribution with m and n degrees of freedom, denoted Fm,n is defined to be
the distribution of the ratio of a chi-squared random variable with degrees of freedom m
divided by m, to an independently distributed chi-squared random variable with degrees
of freedom n, divided by n.
The chi-squared distribution is the distribution of the sum of m squared independent
standard normal random variables. The distribution depends on m, which is called the
degrees of freedom of the chi-squared distribution. A chi-squared distribution with m
degrees of freedom is denoted by χ2m.
5. Let S1, S2, ..., Sn be a sequence of random variables. For example, Sn could be the sample
average Y of a sample of n observations of the random variable Y . The sequence of random
variables {Sn} is said to converge in probability to a limit, µ (that is, Snp−→ µ), if the
probability that Sn is within ±δ of µ tends to 1 as n −→ ∞, as long as the constant δ is
positive. That is, Snp−→ µ if and only if Pr(|Sn − µ| ≥ δ) −→ 0 as n −→∞ for every δ > 0.
If the distributions of a sequence of random variables converge to a limit as n goes to
infinity, then the sequence of random variables is said to converge in distribution.
The central limit theorem says that, under general conditions, the standardized sample
average converges in distribution to a normal random variable. Let F1, F2, ..., Fn be a se-
quence cumulative distribution functions corresponding to a sequence of random variables,
S1, S2, ..., Sn. For example, Sn might be the standardized sample average, (Y − µY )/σY .
Then the sequence of random variables Sn is said to converge in distribution to S (denoted
Snd−→ S) if the distribution functions {Fn} converge to F , the distribution of S. That is,
Snd−→ S if and only if lim
n→∞Fn(t) = F (t) where the limit holds at all points t at which the
limiting distribution F is continous.
11
6. Random sampling is one of the most popular types of random or probability sampling.
In a simple random sample, n objects are selected at random from a population and each
member of the population is equally likely to be included in the sample. The value of
the random variable Y for the ith randomly drawn object is denoted Yi. Because each
object is equally likely to be drawn and the distribution of Yi is the same for all i, the
random variables Y1, ..., Yn are independently and identically distributed (i.i.d.); that is,
the distribution of Yi is the same for all i = 1, ..., n and Y1 is distributed independently
of Y2, ..., Yn and so forth.
7. The sample average or sample mean, Y , of the n observations Y1, ..., Yn is obtained as
follows:
Y =1
n(Y1 + Y2 + ...+ Yn) =
1
n
n∑i=1
Yi
8. (a) Let µY denote some estimator of µY , such as Y or Y1. The estimator µY is unbi-
ased if E(µY ) = µY , where E(µY ) is the mean of the sampling distribution of µY ;
otherwise, µY is biased.
Another desirable property of an estimator µY is that, when the sample size is large,
the uncertainty about the value of µY arising from random variations in the sample is
very small. Stated more precisely, a desirable property of µY is that the probability
that it lies within a small interval of the true value µY approaches 1 as sample size
increases, that is, µY is consistent for µY .
(b) The mean of Y is given by:
E(Y ) =1
n
n∑i=1
E(Yi) = µY .
The variance of Y is given as follows:
σ2Y = var
(1
n
N∑i=1
E(Yi)
)
=1
n2
N∑i=1
var(Yi) +1
n2
N∑i=1
N∑j=1,j 6=i
cov(Yi, Yj)
=σ2Y
n
where Y1, ..., Yn are i.i.d. and Yi and Yj are independently distributed for i 6= j, so
cov(Yi, Yj) = 0.
12
(c) The law of large numbers states that, under general conditions, Y will be near
µY with very high probability when n is large. This is sometimes called the “law
of averages.” When a large number of random variables with the same mean are
averaged together, the large values balance the small values and their sample average
is close to their common mean. The sample average Y converges in probability to
µY (or, equivalently, Y is consistent for µY ) if the probability that Y is in the range
(µY−c) to (µY +c) becomes arbitrarily close to 1 as n increases for any constant c > 0.
The convergence of Y to µ in probability is written, Yp−→ µY . The law of large
numbers says that if Yi, i = 1, .., n are are independently and identically distributed
with E(Yi) = µY and if large outliers are unlikely (technically if var(Yi) = σ2Y <∞,
then Yp−→ µY .
(d) Recall that the mean of Y is µY and its variance is σ2Y
= σ2Y /n. According to the
central limit theorem, when n is large, the distribution of Y is approximately
N(µY , σ2Y
). The distribution of Y is exactly N(µY , σ2Y
) when the sample is drawn
from a population with the normal distribution N(µY , σ2Y ). Therefore, the distri-
bution of the standardized version of Y , (Y − µY )/σY is well approximated by a
standard normal distribution N(0, 1) when n is large.
9. The population regression line is given by Yi = β0 +β1Xi+ui. This is the relationship
that holds between Y and X on average over the population. Thus, if you knew the value
of X, according to this population regression line you would predict that the value of the
dependent variable, Y , is β0 + β1X. Since the coefficients β0 and β1 of the population
regression line are unknown we must use sample data to estimate them. This can be
achieved using the ordinary least squares (OLS) estimators for β0 denoted by β0 and
the OLS estimator for β1 denoted by β1. The OLS regression line, also called sample
regression line, is the straight line constructed using the OLS estimators: β0 + β1X.
The predicted value of Yi given Xi, based on the OLS regression line is Yi = β0 + β1Xi.
The residual for the ith observation is the difference between Yi and its predicted value:
ui = Yi − Yi. Concluding, one can say that the OLS estimators, β0 and β1 are sample
counterparts of the population coefficients, β0 and β1. Similarly, the OLS regression line
β0 + β1X is the sample counterpart of the population regression line β0 + β1X, and the
OLS residuals ui are sample counterparts of the population errors ui.
10. The formula β = (X ′X)−1X ′y calculates the OLS estimators β0 and β1 for our
true unknown population coefficients β0 and β1, where (X ′X) is non-singular matrix.
The OLS estimator minimizes the sum of squared prediction mistakes,∑n
i=1(Yi − b0 −b1X1i−· · ·− bkXki)
2. The above formula for the OLS estimator is obtained by taking the
derivative of the sum of squared prediction mistakes with respect to each element of the
coefficient vector, setting these derivatives to zero, and solving for the estimator β.
13
11. The algebra of OLS in the multivariate model relies on the two symmetric n x n matrices,
PX = X(X ′X)−1X ′ and MX = In − PX . A matrix C is idempotent if C is square
and CC = C. Because PX = PXPX and MX = MXMX and because PX and MX are
symmetric, PX and MX are symmetric idempotent matrices. The matrices PX and MX
can be used to decompose an n-dimensional vector Z into two parts: a part that is
spanned by the columns of X and a part orthogonal to the columns of X. In other words,
PXZ is the projection of Z onto the space spanned by the columns of X, MXZ is the
part of Z orthogonal to the columns of X, and Z = PXZ +MXZ.
12. The least squares estimator is unbiased in every sample. To show this, write
b = (X ′X)−1X ′y = (X ′X)−1X ′(Xβ + ε) = β + (X ′X)−1X ′ε
Now, take expectations, iterating X;
E[b|X] = β + E[(X ′X)−1X ′ε|X]
By Assumption (1), the second term is 0, so
E[b|X] = β
Therefore,
E[b] = EX {[b|X]} = EXβ = β
The interpretation of this result is that for any particular set of observations, X, the least
squares estimator has expectation β. Therefore, when we average this over the possible
values of X we find the unconditional mean is β as well.
To show consistency we write least squares estimator as follows:
b = β +
(X ′X
n
)−1(X ′ε
n
).
If Q−1 exists, then
plim b = β +Q−1plim
(X ′ε
n
)because the inverse is a continuous function of the original matrix. We require the prob-
ability limit of the last term. Let
1
nX ′ε =
1
n
n∑i=1
xiεi =1
n
n∑i=1
wi = w
Then,
plim b = β +Q−1plim w
14
We have Var[w] = E[Var[w|X]] + Var[E[w|X]]. The second term is zero becauseE[εi|xi]=0.
To obtain the first, we use E[ε′iεi|X] = σ2I, so We have Var[w] = E[Var[w|X]] +
Var[E[w|X]]. The second term is zero because E[εi|xi]=0. To obtain the first, we use
E[ε′iεi|X] = σ2I, so
Var[w|X] = E[w′w|X] =1
nX ′E[ε′iεi|X]X
1
n=
(σ2
n
)(X ′X
n
)Therefore,
Var[w] =
(σ2
n
)E
(X ′X
n
)The variance will collapse to zero if the expectation in parentheses is (or converges to) a
constant matrix, so that the leading scalar will dominate the product as n increases. It
then follows that
limn→∞
Var[w] = 0 · Q = 0
Since the mean of w is identically zero and its variance converges to zero, w converges in
mean square to zero, so plim w = 0. Therefore
plimX ′ε
n= 0,
so,
plim b = β +Q−1 · 0 = β
This result establishes that under four Assumptions, b is a consistent estimator of β in
the linear regression model.
Finally, we need to show that the OLS estimator of the population model is asymptot-
ically normally distributed. We start with:
√n(b− β) =
(X ′X
n
)(1√n
)X ′ε
Since the inverse matrix is a continuous function of the original matrix, plim(X ′X/n)−1 =
Q−1. Therefore, if the limiting distribution of the random vector exists, then that limiting
distribution is the same as that of[plim
(X ′X
n
)−1](
1√n
)X ′ε = Q−1
(1√n
)X ′ε
Thus, we must establish the limiting distribution of(1√n
)X ′ε =
√n(w − E[w])
15
where E[w] = 0. We can use the multivariate Lindeberg-Feller version of the central limit
theorem to obtain the limiting distribution of√nw. Using that formulation, w is the
average of n independent random vectors wi = xiεi, with means 0 and variances
Var [x′iεi] = σ2E [x′ixi] = σ2Qi
The variance of√nw is
σ2Qn = σ2
(1
n
)[Q1 +Q2 + · · ·+Qn]
As long as the sum is not dominated by any particular term and the regressors are well
behaved, which in this case, the following holds:
limn→∞
σ2Qn = σ2Q
Therefore, we may apply the Lindeberg-Feller central limit theorem to the vector√nw.
We now have the elements we need for a formal result. If [xiεi], i = 1, . . . . . . , n are
independent vectors distributed with mean 0 and variance σ2Qi <∞ then(1√n
)X ′ε
d−−→ N [0, σ2Q].
It then follows that
Q−1
(1√n
)X ′ε
d−−→ N [Q−10, Q−1(σ2Q)Q−1].
Combining terms,√n(b− β)
d−−→ N [0, σ2Q].
the asymptotic standard normal distribution of b.
13. The standard error of the regression (SER) estimates the standard deviation of the
error term ui. Thus the SER is a measure of the spread of the distribution of Y around
the regression line. In multiple regression, the SER is
SER = su =√s2u where s2
u =1
n− k − 1
n∑i=1
u2i =
SSR
n− k − 1
where SSR is the sum of squared residuals, SSR =∑n
i=1 u2i
The regression R2 is the fraction of the sample variance Yi explained by (or predicted
by) the regressors. Equivalently, the R2 is 1 minus the fraction of the variance of Yi not
explained by the regressors. The mathematical definition of the R2 is the same as for the
16
regression with a single regressor:
R2 =ESS
TSS= 1− SSR
TSS ′
where the explained sum of squares is ESS =∑n
i=1(Yi− Y )2 and the total sum of squares
is TSS =∑n
i=1(Yi − Y )2
Because the R2 increases when a new variable is added to the model, an increase in the R2
when adding an new variable does not necessarily mean that the added variable actually
improves the fit of the model. In this sense, the R2 gives an inflated estimate of how well
the regression fits the data. One way to correct for this is to deflate or reduce the R2 by
some factor, and this is what the adjusted R2 , or R2, does. The adjusted R2, or R2,
is a modified version of the R2 that does not necessarily increase when a new regressor is
added. The R2 is calculated as follows:
R2 = 1− n− 1
n− k − 1
SSR
TSS ′= 1− s2
u
s2Y
The difference between this formula and the one of R2 is that the ratio of the sum of
squared residuals to the total sum of squares is multiplied by the factor (n−1)/(n−k−1).
As the second expression in the formula of R2 shows, this means that the adjusted R2 is
1 minus the ratio of the sample variance of the OLS residuals (with the degree-of-freedom
correction in SER formula) to the sample variance of Y .
14. Recall that PX = X(X ′X)−1X ′, where PX = PXPX and MX = In − PX , where MX =
MXMX . Further recall the OLS predicted values Y = Xβ and the OLS residuals, U =
Y − Y . We now can rewrite Y and U as follows:
Y = Xβ = X(X ′X)−1X ′Y = PXY
U = Y − Y = Y −Xβ = Y −X(X ′X)−1X ′Y = (In −X(X ′X)−1X ′)Y = MXY =
MX(Xβ + U) = MXXβ +MXU = (In −X(X ′X)−1X ′)X +MXU = MXU
Hence, the squared SER, s2u, can be written as:
s2u =
1
n− k − 1
n∑i=1
u2i =
1
n− k − 1U ′U =
1
n− k − 1U ′MXU
where the final equality follows because U ′U = (MXU)′(MXU) = U ′MXMXU = U ′MXU
(because MX is symmetric and idempotent).
17
15. The t-Statistic is a statistic used to perform a hypothesis test. In general the t-statistic
has the form
t =estimator-hypothesized value
standard error of the estimator.
The null and alternative hypothesis need to be stated precisely before they can be tested.
As an example, assume we want to test hypotheses about the slope β1. The null hypothesis
and the two-sided alternative hypothesis are
H0 : β1 = β1,0 vs. H1 : β1 6= β1,0 (two− sided− alternative)
One could also use a one-sided alternative hypothesis here and test against H1 : β1 >
or < β1,0. The first step is to compute the standard error of β1, SE(β1). The standard
error of β1 is an estimator of σβ1 the standard deviation of the sampling distribution of
β1. Specifically,
SE(β1) =√σ2β1
where
σ2β1
=1
n×
1n−2
∑ni=1 (Xi − X)2u2
i
[ 1n
∑ni=1 (Xi − X)2]2
The second step is to compute the t-statistic,
t =β1 − β1,0
SE(β1)
The third step is to compute the p-value, the probability of observing a value of β1 at
least as different from β1,0 as the estimate actually computed (βact1 ), assuming that the
null hypothesis is correct. Stated mathematically,
p-value = PrH0
[∣∣∣β1 − β1,0
∣∣∣ > ∣∣∣βact1 − β1,0
∣∣∣]= PrH0
[∣∣∣∣∣ β1 − β1,0
SE(β1)
∣∣∣∣∣ >∣∣∣∣∣ βact1 − β1,0
SE(β1)
∣∣∣∣∣]
= PrH0(|t| > |tact|)
where PrH0 denotes the probability computed under the null hypothesis. Because β1
is approximately normally distributed in large samples, under the null hypothesis the
t-statistic is approximately distributed as a standard normal random variable, so in large
samples,
p-value = Pr(|Z| > |tact|) = 2Φ(−|tact|)
A p-value of less than 5% provides evidence against the null hypothesis in the sense that,
under the null hypothesis, the probability of obtaining a value of β1 at least as far from the
18
null as that actually observed is less than 5%. If so, the null hypothesis is rejected at the
5% significance level. Alternatively, the hypothesis can be tested at the 5% significance
level simply by comparing the absolute value of the t-statistic to 1.96, the critical value
for a two-sided test, and rejecting the null hypothesis at the 5% level if |tact| > 1.96.
The F-Statistic is used to test joint hypotheses. When the joint null hypothesis has the
two restrictions that β1 = 0 and β2 = 0, the F-statistic combines the two t-statistics t1
and t2 using the formula
F =1
2
(t21 + t22 − 2ρt1,t2t1t2
1− ρ2t1,t2
)where ρt1,t2 is an estimator of the correlation between the two t-statistics. If we want to
test for more restrictions, say q, we need to use the following approach. Consider a joint
hypothesis that is linear in the coefficients and imposes q restrictions, where q ≥ k + 1.
Each of these q restrictions can involve one or more of the regression coefficients. This
joint null hypothesis can be written in matrix notation as
Rβ = r
where R is a q × (k + 1) nonrandom matrix with full row rank and r is nonrandom
q × 1 vector. The number of rows of R is q, which is the number of restrictions being
imposed under the null hypothesis. The heteroskedasticity-robust F-statistic testing the
joint hypothesis in matrix form is
F = (Rβ − r)′[RΣβR′]−1(Rβ − r)/q
16. A 95% two-sided confidence interval for β1 is an interval that contains the true value
of β1 with a 95% probability; that is, it contains the true value of β1 in 95% of all possible
randomly drawn samples. Equivalently, it is the set of values of β1 that cannot be rejected
by a 5% two-sided hypothesis test. When the sample size is large, it is constructed as
95% confidence interval for β1 = [β1 − 1.96SE(β1), β1 + 1.96SE(β1)].
17. The error term ui is homoskedastic if the variance of the conditional distribution of ui
given Xi, var(ui|Xi = x), is constant for i = 1, ..., n and in particular does not depend on
x. Otherwise the error term is heteroskedastic. The homoskedasticity-only standard
error of β1 is SE(β1) =√σ2β1
, where σ2β1
is the homoskedasticity-only estimator of the
19
variance of β1:
σ2β1
=s2u∑n
i=1 (Xi − X)2
The heteroskedasticity-robust standard errors can be obtained via
σ2β1
=1
n×
1n−2
∑ni=1 (Xi − X)2u2
i
[ 1n
∑ni=1 (Xi − X)2]2
Because homoskedasticity is a special case of heteroskedasticity, the estimator σ2β1
of the
variances of β1 given above produces valid statistical inferences whether the errors are
heteroskedastic or homoskedastic. Thus hypothesis tests and confidence intervals based
on those standard errors are valid whether or not the errors are heteroskedastic. This is
not the case for σ2β1
which can only be used under homoskedasticity.
18. If the three least squares assumptions (1. The error term ui has conditional mean zero
given Xi: E(ui|Xi) = 0, 2. (Xi, Yi), i = 1, ..., n, are i.i.d. draws from their joint distribu-
tion, 3. Large outliers are unlikely: Xi and Yi have nonzero finite fourth moments) hold
and if errors are homoskedastic, then the OLS estimator β1 is the Best (most efficient)
Linear conditionally Unbiased Estimator (BLUE).
19. For the following proofs we make use of the following six assumptions:
(1) E(ui|Xi) = 0 (ui has conditional mean zero)
(2) (Xi, Yi), i = 1, ..., n, are independently and identically distributed (i.i.d.) draws from
their joint distribution
(3) Xi and ui have nonzero finite fourth moments.
(4) X has full column rank (there is no perfect multicollinearity)
(5) var(ui|Xi) = σ2u (homoskedasticity)
(6) The conditional distribution of ui given Xi is normal (normal errors)
(a) The first and second assumptions imply that E(ui|X) = E(ui|Xi) = 0 and that
cov(ui, uj|X) = E(uiuj|X) = E(uiuj|Xi, Xj) = E(ui|Xi)E(uj|Xj) = 0 for i 6= j.
The first, second, and fifth assumptions imply that E(u2i |X) = E(u2
i |Xi) = σ2u. Com-
bining these results, we have that under Assumptions (1) and (2), E(U |X) = 0n,
and under Assumptions (1), (2) and (5), E(UU ′|X) = σ2uIn, where 0n is the n-
dimensional vector of zeros and In is the n × n identity matrix. Similarly, the
first, second, fifth, and sixth assumptions imply that the conditional distribution of
the n-dimensional random vector U , conditional on X is the multivariate normal
distribution. That is, under Assumptions (1), (2), (5) and (6), the conditional
distribution of U given X is N(0n, σ2uIn).
20
(b) Because β = β+(X ′X)−1X ′U and because the distribution of U conditional on X is,
by assumption, N(0n, σ2uIn), the conditional distribution of β given X is multivariate
normal with mean β. The covariance matrix of β, conditional on X, is σβ|X = E[(β−β)(β − β)′|X] = E[(X ′X)−1X ′UU ′X(X ′X)−1|X] = (X ′X)−1X ′(σ2
uIn)X(X ′X)−1
= σ2u(X
′X)−1. Accordingly, under all six assumptions, the finite-sample condi-
tional distribution of β given X is β ∼ N(β, σ2u(X
′X)−1).
(c) If all six assumptions hold, then s2u = σ2
u has an exact sampling distribution that is
proportional to a chi-squared distribution with n− k − 1 degrees of freedom:
s2u ∼
σ2u
n− k − 1× χ2
n−k−1
The proof starts with the following equation:
s2u =
1
n− k − 1
n∑i=1
u2i =
1
n− k − 1U ′U =
1
n− k − 1U ′MxU.
Because U is normally distributed conditional on X and because Mx is a symmet-
ric idempotent matrix, the quadratic form U ′MxU/σ2u has an exact chi-squared
distribution with degrees of freedom equal to the rank of Mx. The rank of
Mx is n− k − 1. Thus U ′MxU/σ2u has an exact χ2
n−k−1 distribution, from which the
given equation follows.
(d) If (i) Z has a standard normal distribution, (ii) W has a chi-squared distribution with
m degrees of freedom distribution, and (iii) Z and W are independently distributed,
then the random variable Z/√
(W/m) has a t-distribution with m degrees of freedom.
We start the proof with the following Equation where su = σu:
tj =βj − βj,0
su√
[(X ′X)−1]jj
To put tj in the above mentioned form, we rewrite the equation as
tj =(βj − βj,0)/
√σ2u[(X
′X)−1]jj√W/(n− k − 1)
where W = (n − k − 1)/(s2u/σ
2u), and let Z = (βj − βj,0)/
√σ2u[(X
′X)−1]jj and
m = n−k−1. With these definitions, tj = Z/√W/m. Thus, to prove the statement
given in the question, we must show (i) through (iii) for these definitions of Z, W
and m.
(i) An implication of the results of exercise (b) is that, under the null hypothesis,
Z = (βj − βj,0)/√σ2u[(X
′X)−1]jj has an exact standard normal distribution, which
shows (i).
21
(ii) From exercise (c) we know that W has a chi-squared distribution with n− k− 1
degrees of freedom which shows (ii).
(iii) To show (iii), it must be shown that βj and s2u are independently distributed.
We already know that β − β = (X ′X)−1X ′U and s2u = (MxU)′(MxU)/(n− k − 1).
Thus β−β and s2u are independent if (X ′X)−1X ′U and MxU are independent. Both
(X ′X)−1X ′U and MxU are linear combinations of U , which has a N(0n×1, σ2uIn)
distribution conditional on X. But because MXX(X ′X)−1 = 0n×(k+1), it follows that
(X ′X)−1X ′U and MxU are independently distributed. Consequently under all six
assumptions above βj and s2u are independently distributed, which shows (iii) and
thus completes the proof.
20. Perfect multicollinearity arises when one of the regressors is a perfect linear combination
of the other regressors. In case of perfect multicollinearity it is impossible to compute
the OLS estimator. Imperfect multicollinearity arises when one of the regressors is very
highly correlated - but not perfectly correlated - with the other regressors. Unlike perfect
multicollinearity, imperfect multicollinearity does not prevent estimation of the regression,
nor does it imply a logical problem with the choice of the regressors. However, it does
mean that one or more regression coefficients could be estimated imprecisely.
21. Omitted variable bias is the bias in the OLS estimator that arises when the regressor,
X, is correlated with an omitted variable. For omitted variable bias to occur, two condi-
tions must be met:
(1) X is correlated with the omitted variable.
(2) The omitted variable is a determinant of the dependent variable, Y .
22. Functional form misspecification arises when the functional form of the estimated
regression function differs from the functional form of the population regression function.
If the functional form is misspecified, then the estimator of the partial effect of a change
in one of the variables will, in general, be biased. Fuctional form misspecification often
can be detected by plotting the data and the estimated regression function, and it can be
corrected by using a different functional form.
Errors-in-variables in the OLS estimator arises when an independent variable is mea-
sured imprecisely. This bias depends on the nature of the measurement error and persists
even if the sample size is large. If the measured variable equals the actual value plus a
mean-zero, independently distributed measurement error, then the OLS estimator in a
regression with a single right-hand variable is biased toward zero, and its probability limit
is given by:
β1p−→ σ2
X
σ2X + σ2
w
β1
The measurement error in Y is different from measurement error in X (Error-in-
22
variable). If Y has a classical measurement error, then this measurement error increases
the variance of the regression and of β1S but does not induce bias in β1.
Simultaneous causality bias, also called simultaneous equations bias, arises in a re-
gression of Y on X when, in addition to the causal link of interest from X to Y , there
is a causal link from Y to X. This reverse causality makes X correlated with the error
term in the population regression of interest.
23. Logarithms can be used to transform the dependent variable Y , an independent variable
X, or both (but the variable being transformed must be positive). The following table
summarizes these four cases and the interpretation of the regression coefficient β1. In each
case, β1 can be estimated by applying OLS after taking the logarithm of the dependent
and/or independent variable.
Case Regression Specification Intepretation of β1
linear-linear Yi = β0 + β1Xi + ui A change in X by one unit
(∆X = 1) is associated with
a change in Y of β1.
log-linear ln(Yi) = β0 + β1Xi + ui A change in X by one unit
(∆X = 1) is associated with
a 100β1% change in Y.
linear-log Yi = β0 + β1ln(Xi) + ui A 1% change in X is associated
with a change in Y of 0.01β1.
log-log ln(Yi) = β0 + β1ln(Xi) + ui A 1% change in X is associ-
ated with a β1% change in Y ,
so β1 is the elasticity of Y with
respect to X.
24. Consider the population regression of Yi against two binary (only taking on values 0 or
1) variables D1i and D2i. The population linear regression of Yi on these two variables is
Yi = β0 + β1D1i + β2D2i + ui
The specification of the model has an important limitation: The effect of D1i in this
specification, holding constant D2i, is the same for both possible values of D2i. There is,
however, no reason that this must be so. Phrased mathematically, the effect on Yi of D1i,
holding D2i constant, could depends on the value of D2i. Although this specification does
not allow for this interaction between D1i and D2i, it is easy to modify the specification
so that it does by introducing another regressor, the product of the two binary variables
23
D1i ×D2i. The resulting regression is
Yi = β0 + β1D1i + β2D2i + β3(D1i ×D2i) + ui
The new regressor, the product D1i × D2i, is called interaction term or interacted
regressor. The interaction term allows the population effect on Yi of Changing D1i (from
D1i = 0 to D1i = 1) to depend on D2i.
24
Statistical Questions
1. Assume that the random variable X follows an unknown distribution with E(X) = 2 and
V ar(X) = 4. A random variable Y also follows an unknown distribution with E(Y ) = −2,
and V ar(Y ) = 9. Additionally, ρXY = 0.5. Now consider the random variable Z with:
Z = ωX + (1− ω)Y, ω ∈ (0, 1]
(a) Calculate the expected value and the variance of Z.
(b) Determine the value of ω such that the variance of Z is minimized.
(c) Now assume that ρXY = 0. What consequences for the dependence between Y and
X do you suspect? What follows if X and Y follow a Normal distribution?
2. Assume that the continuous random variable has an expected value of E(X) = 2 and
variance V ar(x) = 1. What conclusions do you draw about the probability P (0 < X <
4)?
3. Consider the random variable Y having probability density function:
fY (y) = 32(y − 0.5y2), 0 < y < 2
(a) Calculate the expected value and the variance of Y .
(b) Now consider the second random variable Z having the same probability density
function as Y . Correlation between Y and X is given by 3V ar(Y ). Calculate the
variance of the random variable T given by T = 2Y + Z
4. Consider an urn containing N = 10 balls, of which M = 3 are red. From this urn you
draw n = 3 times one ball without replacement. Calculate the probability of at least
drawing 2 red balls.
5. Consider an identically and independently distributed random sample, (x1, ..., xn) , from
an exponentially-distributed population having probability density function f(x) = 1θe−
xθ .
(a) Set up the log likelihood function and calculate the Maximum-Likelihood-Estimator
θML for θ.
(b) Check for biasedness of the ML-estimator and calculate the Mean-Squared-Error
(MSE). Is θML MSE-consistent?
6. Consider the following results of a horse race:
Horse A B C D∑
Number of wins 10 16 12 22 60
25
(a) For a significance level of α = 0.05 check the hypothesis, whether the number of
wins is uniformly distributed across the 4 horses.
(b) Now consider the number of wins of horse D. For a significance level of α = 0.01
check the hypothesis of a Bernoulli-distribution with parameter ρ = 1/3 (d = 1, if
horse D won the race, d = 0 otherwise)
(c) Calculate the probability of horse D winning at most twice in 10 races under the
assumption of a Bernoulli-distribution from part b) with ρ = 1/3
7. The random variable X has the following probability density function:
f(x)=
4/x5 if x > 1
0 else
(a) Determine the first order non-central moment and the second order central moment.
(b) Determine the probability density function of the random variable Z = ln(X2)
8. Entrepreneur Meyer produces wooden slats whose lengths are to be of 1 meter. He wants
to increase productivity by installing a new machine. In order to control for the quality
of the slats from the new machine, n1 = 101 randomly chosen slats are measured. The
mean length is 99cm and the sample variance is 30 cm2. Assume the wooden slats to be
normally distributed.
(a) Test the null hypothesis that the mean length coincides with the scheduled value.
Calculate the p-value of the conducted test. Assume a significance value of α = 0.05
(b) Now assume that the entrepreneur wants to be statistically sure that the mean length
of the wooden slats produced by the new machine undercuts 100cm. Formulate the
null and the alternative hypotheses, conduct the test and calculate the p-value.
Hint: Assume a large sample size.
9. A student has no time to prepare for a multiple choice test of 20 questions. He decides
to have a guess at all the questions. Each question has 5 possible answers.
(a) How is the random variable distributed, that gives the number of correct answers?
(b) How many questions will the student answer correctly on average?
(c) The test is passed if 10 questions are answered correctly. What is the probability of
the student to pass the test?
(d) How should the threshold for passing the test be defined, if the student’s chance to
pass the test by guessing the answers, should be larger than 5%?
26
10. After the game ended in a tie the penalty shoot-out takes place. Each team has 5 shoots
and the team which scores more often than the other wins. Assume that the single shoots
are independent from each other and each player scores with a probability of 0.8. What
is the probability that after 10 shots (5 shots per team) the game is decided?
11. Which distributions do the following random variables X possess?
(a) X1= Number of items of a rarely sold product demanded in one day
(b) X2= Time passing between two worst-case scenarios in nuclear energy generation
(c) X3= Number of telephone calls in a large call centre within one hour
(d) X4= Number of correct guesses in the lottery ”6 out of 49”
12. Assume your little sister does handicrafts and the necklace consists of 50 parts which have
on average a length of 2 cm and a standard deviation of 0.2 cm. Which distribution does
the length of the whole necklace follow?
13. A bank knows that the mean indebtedness of its student customers at the end of the
academic year is distributed approximately normally with a mean of 300 and a standard
deviation of 100. What is the probability that a student chosen at random will have a
debt:
(a) Less than 100
(b) Between 100 and 200
(c) Over 450
(d) A debt of 400
14. A dating agency usually claims in its advertising that the mean age of the members on its
list is lower than 40. As a check, it takes a sample of 30 members and obtains a sample
mean age of 37 years and a sample variance of 25 years. Test whether its claim is justified
with:
(a) α = 0.05
(b) α = 0.01
15. A trade union believes in the basis of its membership records that at least four out of
every five union members in the financial services sector are members of its union. A
smaller rival union denies this statement and wants to show that the proportion is in fact
smaller than the one claimed by the larger union. It finds that the proportion belonging
to the larger rival union is 0.72. Conduct the associated test using
(a) α = 0.05
(b) α = 0.01
27
Statistical Questions: RESULTS
1. a) E(Z) = 4ω − 2; V AR(Z) = 7ω2 − 12ω + 9
b) ω = 6/7
c) ρ = 0 implies that there is no linear relationship among x and y. If x and y follow a
bivariate normal distribution, ρ = 0 implies that x and y are independent.
Z ∼ N(4ω − 2, 13ω2 − 18ω + 9)
2. P (0 < X < 4) = 0.75
3. a) E(Y ) = 1; V AR(Y ) = 0.2
b) V AR(T ) = 1.48
4. P (X ≥ 2) = 11/60
5. a) ln(L) = n ln(1/θ)− 1/θ∑
iXi; θML = x
b) the estimator is unbiased and MSE-consistent
c) weglassen
6. a) reject H0, accept H1
b) don’t reject H0
c) P (X ≤ 2) = 0.299
7. E(X) = 1.333; E(X2) = 2→ E[(X − µ)2] = 0.223
8. a) don’t reject H0; pvalue=0.0702
b) reject H0, accept H1; pvalue=0.0333
9. a) X ∼ Bin(n = 20, p = 1/5)
b) E(X) = 4
c) P (’pass the test’) = 0.0026
d) the threshold should be ’at least 6 correct answers’.
10. P (’decision’) = 0.8855
11. a) poisson distribution
b) exponential distribution
c) normal distribution
d) hypergeometric distribution.
12. N(100 cm; 2 cm2)
13. a) P (X < 100) = 0.0228
b) P (100 < X < 200) = 0.1359
c) P (X > 450) = 0.0668
d) P (X = 450) = 0
28
14. a) reject H0, accept H1
b) reject H0, accept H1
15. a) reject H0, accept H1
b) don’t reject H0
29
Macroeconomic Questions
Multiple-Choice Questions
1. The effectiveness of monetary policy depends on the LM-curve and does not depend on
the IS-curve.
(a) True
(b) False
2. The effectiveness of fiscal policy depends on the IS-curve and does not depend on the
LM-curve.
(a) True
(b) False
3. An expansionary fiscal policy raises the trade surplus.
(a) True
(b) False
4. Higher inflationary expectations decrease the nominal interest rate.
(a) True
(b) False
5. On the long run the growth in money supply significantly affects the price levels.
(a) True
(b) False
6. A fast adjustment on the money market implies that the economy always moves along
(a) The IS-curve
(b) The LM-curve
(c) The IS- as well as the LM-curve
(d) Neither the IS- nor the LM curve
7. An increase of government demand raises the
(a) Interest rates
(b) Investments
(c) Interest rates and investment
30
(d) Neither interest rates nor investment
8. A high interest rate sensitivity of money demand raises
(a) The multiplier of fiscal policy
(b) The multiplier of monetary policy
(c) The multipliers of fiscal as well as monetary policy
(d) Neither the multiplier of fiscal nor of monetary policy
9. Compared to a closed economy the simple multiplier of an open economy is
(a) Larger
(b) Smaller
(c) The same
(d) a), b) and c) hold
10. An increase in government spending causes an increase in investment
(a) In the short run
(b) In the long run
(c) In the short as well as In the long run
(d) Neither in the short nor in the long run
11. Unfavorable supply shocks
(a) Increase prices and output
(b) Reduce prices and increase output
(c) Increase prices and reduce output
(d) Reduce prices and output
Goods market
Consider the following goods market model
Y = Y s = Y d
Y d = C + Ia +G
C = a+ b · Y v (a > 0, 0 < b < 1)
Y v = Y − T
T = t · Y (0 < t < 1)
31
where Y = income, Y s = goods supply, Y d = goods demand, Y v = disposable income, C =
consumption, Ia = exogenous/autonomous investments, G = government spending, T = taxes,
t = tax rate.
1. Derive the equilibrium level of income.
2. Derive the investment and government spending multiplier (dY/dI and dY/dG).
3. Why does the multiplier process converge?
4. Derive the government spending multiplier in case the government spending is completely
tax financed (dG = dT ). Compare your result to the multiplier derived in exercise 2.
IS/LM model
Consider the following IS/LM model for the closed economy (in reduced form):
Y = C((1− t)Y ) + I(i) +G
M/P = L(Y, i)
where Y = income, C = consumption, I = investment, G = government spending, M = money
supply, L = money demand, i = interest rate, P = fixed price level.
1. Analyze graphically (in a i/Y diagram) the effects of an increase in
(a) government spending (dG > 0).
(b) the money supply (dM > 0).
Describe verbally the adjustment process.
2. Derive the government spending multiplier. Why is it smaller than (or equal to) the
elementary multiplier with fixed investments? For which special cases is it equal to the
elementary multiplier?
IS/LM/Z (Mundell-Fleming) model
Consider the IS/LM/Z model of the small open economy:
Y = C((1− t)Y ) + I(i) +G+ A(Y, Ya, e)
M
P= L(Y, i)
Z = 0 = An(Y, Ya, e) +K(i, ia)
32
where Y = (domestic) income, C = consumption, I = investment, G = government spending,
M = money supply, L = money demand, i = interest rate, ia = foreign interest rate, P =
fixed price level, A = net exports, Ya = foreign income, Z = balance of payments surplus, An
= current account surplus, K = capital account surplus, e = exchange rate.
1. Why is a current account surplus associated with a capital account deficit in the case of
flexible exchange rates? Why is this not necessarily the case under fixed exchange rates?
2. Assume flexible exchange rates. Analyze verbally and graphically (in a i/Y diagram) the
effects of an increase in
(a) government spending (dG > 0).
(b) the money supply (dM > 0).
3. Show graphically that under flexible exchange rates and in the limit case of perfect capital
mobility, fiscal policy is inefficient.
4. Why is monetary policy inefficient in case of a fixed exchange rate regime?
AD/AS model
Consider the flex-price model of a closed economy with wage rigidity:
N = Nd(W/P ) < N s(W/P )
Y = Y s = Y (N,K)
Y d = C((1− t)Y ) + I(i) +G
M/P = L(Y, i)
where Y = income, C = consumption, I = investment, G = government spending, M = money
supply, L = money demand, i = interest rate, P = flexible price level, N = employed labor,
Nd = labor demand, N s = labor supply, W = wage, K = fixed capital stock, t = income tax
rate. Wages are inflexible downwards.
1. Explain the Keynes effect.
2. Analyze verbally and graphically (in a P/Y diagram) the effects of an increase in
(a) government spending (dG > 0).
(b) the money supply (dM > 0).
3. Assume that wages are completely flexible (Classical version) such that N = Nd(W/P ) =
N s(W/P ). Show that fiscal and monetary policies are inefficient with respect to output
in this case.
33
(Static) New Keynesian Macroeconomics
Consider the static approximation of the baseline New Keynesian model:
x = a− b · r + ε1
x = y − y
r = i− πe
i = rn + πT + kπ(π − πT ) + kx · x
rn =a
b
π = πe + δ · x+ ε2
where x = output gap, y = output, y = flex-price output, r = real interest rate, i = nominal
interest rate, πe = inflation expectations, rn = natural real interest rate, πT = inflation target
level of the central bank, ε1 = demand shock, ε2 = supply shock. Assume that the inflation
expectations are equal to the targeting rule of the central bank.
1. Explain the Taylor principle in the interest rule of Taylor-type.
2. Name the differences to the IS/LM-AS/AD model.
3. Analyze verbally and graphically (in a π/x diagram) the effects of a
(a) positive demand shock.
(b) positive supply shock.
(c) anticipated and credible increase in the target level of the central bank (πe = πT ).
(d) unanticipated increase in the target level (dπT > 0, dπe = 0).
Dynamic overshooting model of Dornbusch-type
Consider the dynamic Dornbusch-type model of a small open economy:
y = (a0 + a1y − a2(i− E(p))) + g + (b0 − b1y + b2y∗ − b3τ)
τ = p− (p∗ + e)
m− p = l0 + l1y − l2i
i = i∗ + E(e)
p = π + δ(y − y)
with: y = real income or real output, i = nominal interest rate, i− p = real interest rate, p =
inflation rate, g = government expenditure, τ = terms of trade, e = flexible nominal exchange
rate in price notation, e = rate of change of the exchange rate, m = money supply (exogenously
34
given), m = growth rate of money supply (control variable of the central bank), y = natural
or long-term output level, π = augmentation term of the Phillips curve. Foreign variables
(y∗, i∗, p∗) are denoted by a superscript star. A dot above a variable stands for the derivative
of this variable with respect to time t, e.g. p = dp/dt.
The model can be reduced into a two-dimensional system of the form(a2 λ/δ
l2 l2 − l1/δ
)(τ
mr
)=
(b3 0
0 1
)(τ
mr
)+
(λ · y − a0 − b0 − g − b2 · y∗ + a2(i∗ − p∗)−l0 − l1 · y + l2(i∗ − p∗) + l2 · m
)
with λ = 1 − a1 + b1 > 0 and mr = m − p. Assume that l2 < l1/δ, implying that the system
matrix has one stable and one unstable root. In a τ/mr-diagram, the stable saddle arm has a
negative slope and the unstable arm has a positive slope.
1. Assume that the terms of trade are forward-looking and the real money stock is prede-
termined. Why does this assumption ensure the uniqueness and stability of the model
solution?
2. Compute the steady state values of τ and mr.
3. Analyze graphically in a τ/mr-diagram the effects of dm = 1.
Dynamic New Keynesian Macroeconomics – Microfoundation
The representative household seeks to maximize the objective function
Et
∞∑k=0
βkU(Ct+k, Nt+k)
with period utility function
U(Ct, Nt) =C1−σt
1− σ− N1+η
t
1 + η
where Ct is the quantity consumed, Nt is the hours of work, Pt is the price of the consumption
good. Each period the household faces the following budget constraint
(1 + it−1)Bt−1 +WtNt − T nt = PtCt +Bt
where Wt is the nominal wage, Bt is the quantity of one-period, nominally riskless bonds, it is
the corresponding nominal interest rate and T n captures lump-sum taxes and dividends.
1. Derive the Euler equation by solving the household’s optimization problem.
2. Log-linearize the Euler equation around the steady state.
35
Macroeconomic Questions: RESULTS
Goods market
1. Y0 = 11−b(1−t)(a+ Ia +G)
2. dYdIa
= dYdG
= 11−b(1−t)
3. Due to taxes and savings.
4. dYdG
∣∣dT=dG
= 1 (Haavelmo-Theorem)
IS/LM model
1. (a) G ↑→ Y ↑, i ↑
(b) M ↑→ Y ↑, i ↓
2. dYdG
= 11−CY v (1−t)−IiLy/Li . Ii = 0 or Ly → −∞ (liquidity trap) give the elementary multi-
plier of the goods market model.
IS/LM/Z (Mundell-Fleming) model
1. e flexible implies Z = 0 and thus An(Y, Ya, e) = −K(i, ia).
2. (a) G ↑→ Y ↑, i ↑, e ↑ (if capital mobility is sufficiently small)
G ↑→ Y ↑, i ↑, e ↓ (if capital mobility is sufficiently large, but not perfect)
(b) M ↑→ Y ↑, i ↓, e ↑
3. Under perfect capital mobility: dA = −dG (complete crowding out).
4. Under fixed exchange rate: Decrease in reserves completely neutralizes the effects of the
expansionary monetary policy.
AD/AS model
1. P ↓→ MP↑→ i ↓→ Y ↑
2. (a) G ↑→ Y ↑, i ↑, P ↑
(b) M ↑→ Y ↑, i ↓, P ↑
3. Monetary policy: dM = dP = dW → dN = 0→ dY = 0.
Fiscal policy: dG = −dI → dY = 0 (investment crowding out).
36
(Static) New Keynesian Macroeconomics
1. kπ > 1.
2. Most important differences: Goods demand depends on real interest rate. Control vari-
ables of the central banks is the nominal interest rate. Supply side is described by an
inflation equation (Phillips curve), which depends on expected future inflation.
3. (a) ε1 ↑→ x ↑, π ↑, i ↑
(b) ε1 ↑→ x ↓, π ↑, i ↑
(c) πT = πe ↑→ dx = 0, π ↑, i ↑
(d) πT ↑ (dπe = 0)→ dx ↑, π ↑, i ↓
Dynamic overshooting model of Dornbusch-type
1. Since system matrix has one stable and one unstable eigenvalue (cf. Blanchard-Kahn-
conditions).
2. (τ
mr
)= −
(1b3
[λ · y − a0 − b0 − g − b2 · y∗ + a2(i∗ − p∗)]−l0 − l1 · y + l2(i∗ − p∗) + l2 · m
)
3. dτ = 0, dm < 0. τ ≥ 0, mr ≤ 0 ∀ t > 0.
Dynamic New Keynesian Macroeconomics – Microfoundation
1. C−σt = β(1 + it)Et
(PtPt+1
)EtC
−σt+1
2. ct = Etct+1− 1σ(it−Etπt+1 + log β) with πt = logPt− logPt−1 and ct = logCt− logC. C
is the steady state of Ct.
37
Microeconomic Questions
Firm Theory
1. Define the concept of a production function.
2. Define the concept of an isoquant.
3. What is the marginal product of a particular input?
4. Define the concept of increasing (constant, decreasing) returns to scale.
5. Consider the following production function
f(x1, x2) = xa1xb2, a, b > 0
(a) Determine the marginal product of each factor.
(b) Determine the technical rate of substitution of input 1 for input 2.
(c) Show that the technical rate of substitution equals the slope of the isoquant.
(d) Under what circumstances does this production function exhibit increasing, constant
or decreasing returns to scale?
(e) Let the output be denoted by y, the output price by p. Input prices for factors 1
and 2 are given by w1 and w2, respectively. Write down the profit equation.
6. A firm produces an output good q using one input good x according to the production
function f(x) = 4√x. The output price is given by p, the input price is given by w.
Determine the profit maximizing factor demand function and the supply function.
7. What does it mean if two production factors are perfect substitutes (perfect comple-
ments)?
8. Consider the following production functions. For which functions are the production
factors perfect substitutes or perfect complements?
(a) f(x1, x2) = x1 + x2
(b) f(x1, x2) =√x1 +
√x2
(c) f(x1, x2) =√ax1 + bx2
(d) f(x1, x2) = min{a√x1, b√x2}
(e) f(x1, x2) = [xρ1 + xρ2]1ρ
9. Define the concept of a profit function and of a cost function.
38
Consumer Theory
1. Explain what it means if a utility function represents a certain preference ordering. Ex-
plain the concept of an indifference curve.
2. Sweetie only consumes chocolate bars and lollies. The amounts of chocolate bars and
lollies are given by xC and xL, respectively. Her preferences can be represented by the
following utility function:
U(xC , xL) = ln(xC) + ln(xL)
Prices for chocolate bars and lollies are given by pC and pL, respectively. Sweetie’s income
is denoted by m.
(a) Write down her budget constraint.
(b) Determine the utility maximizing consumption bundle.
(c) Assume prices are given by pC = pL = 2. Sweetie’s income is given by m=500. How
many chocolate bars and how many lollies does Sweetie consume in the optimum?
(d) Assume the price of chocolate bars decreases to p′C = 1. Determine the new con-
sumption bundle. Explain why the change in optimal consumption (the total effect
of the price change) can be decomposed into a substitution effect and an income
effect
3. Oskar’s preferences for pears and apples can be represented by the following utility func-
tion:
U(x1, x2) = 2x1 + x2
where x1 denotes the amount of pears and x2 the amount of apples he consumes. The
price of a pear is given by p1 = 0.5, the price of an apple is given by p2 = 0.2.
(a) How many pears and apples will Oskar consume if he has 3 Euros?
(b) Determine the marginal rate of substitution of pears for apples and interpret your
result.
4. Explain the difference between a normal and an inferior good.
39
Markets
1. Assume the demand for bikes is given by
QD = 50− 32P
Supply is given by
QS = 12P .
(a) Determine equilibrium price and quantity.
(b) Determine the price elasticity of supply and demand in equilibrium and interpret
your results.
2. Assume the demand for strawberries (in kg) in Schleswig-Holstein is given by
QD = 2000− 300P .
Supply is given by
QS = 200P .
(a) Determine equilibrium price and quantity.
(b) Assume a subsidy s=1 per kilogram strawberries is introduced. Determine the new
equilibrium (i.e. quantity, producers’ price P S and consumers’ price P c).
3. A monopolist with cost function C(y) = 6y + 28 is facing the inverse demand function
P (y) = 30 − 2y. Here, y denotes his level of output. Determine monopoly price and
quantity, profit of the monopolist as well as producers’ and consumers’ surplus.
40
Microeconomic Questions: RESULTS
Firm Theory
1. We may look at a firm that produces one output y using two inputs. The production
function f(x1, x2), with x1, x2 ≥ 0, measures the maximum possible output for given
input quantities x1, x2.
2. The isoquant is the set of all input bundles that produce exactly y units of output.
3. The marginal product of input 1 is the additional amount of output we get if we increase
input 1 by a marginal unit while keeping input 2 constant:
MP1 =∂f(x1, x2)
x1
≥ 0
4. What happens to output if we scale the amount of all inputs up by some constant t?
• Constant returns to scale:
f(t · x1, t · x2) = t · f(x1, x2) = t · y ∀ t ≥ 0 (1)
(If we double all inputs, output also doubles.)
• Increasing returns to scale:
f(t · x1, t · x2) > t · f(x1, x2) = t · y ∀ t ≥ 1 (2)
(If we double all inputs, output more than doubles.)
• Decreasing returns to scale:
f(t · x1, t · x2) < t · f(x1, x2) = t · y ∀ t ≥ 1 (3)
(If we double all inputs, output less than doubles.)
5. (a) MP1 = axa−11 xb2,MP2 = bxa1x
b−12
(b) TRS1,2 = −ax2bx1
(c) We look at a vector of (small) input changes (dx1, dx2). The resulting change in
output is approximated by:
dy =∂f(x1, x2))
∂x1
dx1 +∂f(x1, x2)
∂x2
dx2
41
Along an isoquant, output stays constant:
0 =∂f(x1, x2)
∂x1
dx1 +∂f(x1, x2)
∂x2
dx2
⇒ dx2
dx1
= −∂f(x1,x2))
∂x1∂f(x1,x2))
∂x2
= TRS1,2
(d) a+ b > 1, a+ b = 1, a+ b < 1, respectively.
(e) Π = py − w1x1 − w2x2 = pxa1xb2 − w1x1 − w2x2
6. x(p, w) = 4p2
w2 , y(p, w) = 8pw
7. Perfect complements are input goods that are always used together in fixed proportions.
For example, one needs always two tires and one frame to produce one bicycle.
Two input goods are perfect substitutes if the firm is able to substitute one good for the
other at a constant rate.
8. (a) Perfect substitutes
(b) Neither nor
(c) Perfect substitutes
(d) Perfect complements
(e) Neither nor
9. The profit function yields maximal profits for any given vector of input and output prices:
Π(p, w1, w2).
The cost function measures the minimum cost of producing a given output level for some
fixed input prices: C(y, w1, w2)
Consumer Theory
1. A utility function represents a preference ordering if it assigns a higher number to more-
preferred bundles than to less-preferred bundles:
(x1, x2) � (y1, y2)⇔ u(x1, x2) ≥ u(y1, y2).
An indifference curve is the set of bundles of goods that yield a certain utility level U .
Thus, the consumer is indifferent between all the bundles of this set.
2. (a) m = pCxC + pLxL
(b) x∗C = 12mpC, x∗L = 1
2mpL
(c) She consumes 125 chocolate bars and 125 lollies.
42
(d) x′C = 250, x
′L = 125
Relative prices change such that chocolate bars become relatively cheaper (substi-
tution effect) and real income increases (income effect).
3. (a) Oskar consumes 15 apples and no pear.
(b) MRS1,2 = 2. Oskar is willing to give up 2 apples for one additional pear. Apples
and pears are perfect substitutes.
4. How does a consumer’s demand for a good change as his income changes?
If good i is a normal good, then the demand for it increases when income increases:∂xi∂m
> 0.
If good j is an inferior good, then the demand for it decreases when income increases:∂xj∂m
< 0
Markets
1. (a) P ∗ = 25, Q∗ = 12.5
(b) Demand elasticity: εQD,P = −3. In equilibrium, an increase in the price of one per
cent leads to a decrease in demand of three per cent.
Supply elasticity: εQS ,P = 1. In equilibrium, an increase in the price of one per cent
leads to an increase in supply of one per cent.
2. (a) P ∗ = 4, Q∗ = 800
(b) P c = 3.6, P S = 4.6, Q∗ = 920
3. Price: pM = 18, Quantity: yM = 6, Profit: ΠM = 44,
Producer’s surplus: PSM = 72, CSM = 36
43
Recommended literature:
Students interested in applying for the Master’s degree programme of Kiel University should
dispose of knowledge in the fields of microeconomics, macroeconomics, mathematics, statistics,
and econonometrics comparable to the contents of the following literature (or comparable work):
Microeconomics:
• Varian, H. R. (2010): Intermediate Microeconomics: A Modern Approach, 8th edition,
W. W. Norton & Company, New York.
• Pindyck, R. S., Rubinfeld, D.L. (2005): Microeconomics, 6th edition, Pearson Prentice
Hall, Upper Saddle River, N.J.
Macroeconomics:
• Blanchard, O., Macroeconomics, Prentice Hall
• Mankiw, N., Macroeconomics, Worth Publishers
• Sydsæter, K. et al., Further Mathematics for Economic Analysis, Prentice Hall
Mathematics:
• Sydsæter, K., P. Hammond, Essential Mathematics for Economic Analysis, Prentice Hall
Statistics:
• Mc Clave, J., P. Benson, T. Sincich, Statistics for Business and Economics, Pearson
Econometrics:
• J.H. Stock and M.M. Watson (2012) Introduction to Econometrics, 3.Ed., Pearson (In-
ternational Edition), Chapters 1-9,17, and 17.1-5.
44